mm-2529 === Subject: Re: Easy proof of mathematician lies [...] > So when I say factors of 5, I'm NOT saying multiples of 5. For > instance, 2 and 3 are factors of 6. So when I say factors of 6 it > doesn't mean multiples of 6. And I emphasize that there is a > mathematical term multiples of which applies. [...] In the integers, 5 has the property that if m*n = 5, where m and n are integers, then either m is a unit, or n is a unit. In the algebraic integers, 5 can be further broken down as 5= sqrt(5)*sqrt(5), or as 5 = (2+i)*(2-i) . None of sqrt(5), 2+i or 2-i are units, right? I'd like to know your take on this: Can 5 be broken down into a product of non-unit factors, each of which cannot be further broken down? If so, how? If not, why? Is there a paradox here? Etc. This is an open-ended question... David Bernier === Subject: Re: Easy proof of mathematician lies > Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor > of 5. Does it go with r_1 or r_2? > using standard mathematical usage, while several posters apparently > keep reading factors of 5 as multiples of 5. > The problem has to do with their misuse of mathematical terminology. Do you have a problem with that? There is nothing non-standard here. === Subject: Re: Easy proof of mathematician lies > > I'm putting this bit first to highlight it: >>Note: Keith Ramsay is pointing out that he has edited my post in >>terms of placement of text, as he has moved a section. >> > [...] > | Consider that in the ring of algebraic integers, 5 has algebraic > | integer factors, and given algebraic integers r_1 and r_2, where their > | product is 5, why are you acting as if it's so difficult to comprehend > | that there must be some distribution of factors of 5? > > I'm not acting as if I'm having any kind of difficulty. I'm saying > that *you're* having difficulty in realizing that there must be some > distribution of factors of 5 is a meaninglessly vague phrase. You're > saying this as if to say, Shouldn't this be true?, and not as if > you actually had proven it was true. >>Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in >>some way between r_1 and r_2. >> It makes sense, but (assuming I know what you mean - yes you >> _are_ using the language in nonstandard ways) it's simply >> _not_ _true_. >> Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor >> of 5. Does it go with r_1 or r_2? >using standard mathematical usage, while several posters apparently >keep reading factors of 5 as multiples of 5. Huh? You said that factors of 5 distribute in some way between r_1 and r_2. Take r_1 and r_2 as above and tell me how sqrt(5) distributes among them. >The problem has to do with their misuse of mathematical terminology. That's certainly one of the problems here. But it's _your_ misuse of standard terminology that's the problem. You conveniently ignored the rest of my post where I explained this. >For instance 2 is a factor of 6, but so is 3, and in fact, so is 6. >When you say, factor of, it means something that is a factor of the >given number. And similarly factors of would mean factors of the >given number. Nobody's disputed that. It does not follow that for instance 9 has a factor of 12 - the way the language _is_ used, 9 has a factor of 12 does not mean that there is a factor of 12 which divides 9, which is what you seem to mean, it means that 12 is a factor of 9, which is clearly false. >But you have these posters, like David Ullrich who is a math professor >at Oklahoma State University, who are lost with rather basic >mathematical language to the point that they *keep* arguing *after* >I've corrected them. You see their *belief* system apparently is that >as mathematicians they can't be the ones with the error, so amazingly, >they simply keep almost mindlessly repeating it. >You may guess that they say factors of when they mean multiples >of, but I'm using the proper terminology, in the correct way. >So when I say factors of 5, I'm NOT saying multiples of 5. For >instance, 2 and 3 are factors of 6. So when I say factors of 6 it >doesn't mean multiples of 6. And I emphasize that there is a >mathematical term multiples of which applies. >And in fact, using factors of when you mean multiples of while >common, is technically incorrect. I have never seen anyone, here or elsewhere, use factors of to mean multiples of. Except for that you have a good point. >But people like Ramsay and Ullrich are unlikely to reply to *correct* >their mistakes because the society of sci.math lets people it >considers part of the society get away with the dumbest mistakes. >That's how mathematicians operate as demonstrated before your eyes. >And that's how they can have errors in their discipline for years, and >years because they seek to by definition have a society that is >perfect, when in fact, mathematicians are just people, and people make >mistakes. > ************************ David C. Ullrich === Subject: Re: Easy proof of mathematician lies > Nope. There is no claim that y is a factor as it's *provably* a > factor. > *sigh* Could you give me your definition of a factor, then? I suspect > it's different from what mathematicians usually mean, because the way to > prove y is a factor of x is to find some z such that y*z=x, with suitable > properties for all three of them (i.e. all in the algebraic integers, > say). If you haven't done this, how can you prove it's a factor? I've switched from saying provably a factor, as that is problematic to saying that it's implied to be a factor, and an analogy is in the ring of evens where you have 2 and 6, but 6 does not have 2 as a factor, though 6=2(3), in the ring of integers. Here the problem is that members that should be in the ring of algebraic integers are left out. So imagine one such member 'm', and the algebraic integers 'c' and 'a', such that c = am and you'll find that in the ring of algebraic integers, 'a' is coprime to 'c', which is the problem, as by the definition of factor, 'm' as it's not an algebraic integer is NOT a factor of 'c' in the ring of algebraic integers, just like 2 is NOT a factor of 6 in the ring of evens. > Rather than deal with the term factor as people get confused, in my > paper Advanced Polynomial Factorization, I have three numbers a_1, > a_2, and a_3, and I prove that a_3 is coprime to a factor I call f. > Argh, you profess to want to get away from the term 'factor', and then > you go and use it right again. Please define your terms rigorously. I'm using the standard definition of factor. Here probably I could have said that I prove that a_3 is coprime to an algebraic integer I call f. In the past, I've done the equivalent of saying that 6 in the ring of evens has 2 as a factor, which is incorrect, as the implication is that I'm talking about the ring of evens, when to be correct I have to have switched rings. While 6 does have 2 as a factor in the ring of integers, it does NOT in the ring of evens. What I'm doing now is correcting that mistaken usage, and I'm not surprised that it's taken some time and that many of you may have become confused. The situation, however, is analogous to someone talking about 6 in the ring of evens, saying that 6 has 2 as a factor, when they've had to switch rings. The short answer is that in the ring of evens saying 6 has 2 as a factor is wrong. > Consider c=ab, where 'c' is an algebraic integer, and 'a' is an > algebraic integer, but 'b' is not. > > Now if you include fractions or move to a field like algebraic numbers > that's ok. But I'm talking about 'b' that's not in any way a fraction > or fractional. > It's like with the ring of evens, and given 6 = 2(3), you have that 3 > is outside the ring, as in the ring of evens, 2 is NOT a factor of 6. > > The situation is analogous. > > Do you understand? > No, I don't. You seem to be operating under the assumption that all > are you claiming that the set of evens is incomplete, because 2 > should be a factor of 6, but isn't? Nope. Ok, so you had reason to be confused by my switching around before when I did the equivalent of talk about 2 as a factor of 6 in the ring of evens, when it's not. The problem with algebraic integers is that ring operations lead to a number which *should* be an algebraic integer, but it's not. Here's basically my approach. I get an expression like g = r + fc where g, r, f, and c are algebraic integers. Now I find out that g is not coprime to f, and I can separate off some factor of f, which gives me h = s + c where h appears to be a factor of g, and s appears to be a factor of r. factor of f that is f, was separated off. Now to the appears part, as in checking r, I find that r is NOT an algebraic integer! Astute readers now would probably want to go back to the proof that g is not coprime to f. > How is your situation with the algebraic integers distinct from this, > if it is distinct? I've used the ring of evens because it's simple and keeps people from thinking about fractions. The problem not surprisingly requires that you are very *strict* in your mathematics, and I didn't help with what I explained above in terms of that f word. Still for those of you who prize mathematical knowledge, it's a fascinating little problem, and can even be a challenge to understand, which should only intrigue you more. === Subject: Re: Easy proof of mathematician lies > When you say, factor of, it means something that is a factor of the > given number. And similarly factors of would mean factors of the > given number. Nice of you to clear this up with such an unambiguous statement. We sure don't need circular definitions. -- What goes around, comes around. -- Unknown -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Easy proof of mathematician lies > In the past, I've done the equivalent of saying that 6 in the ring of > evens has 2 as a factor, which is incorrect, as the implication is > that I'm talking about the ring of evens, when to be correct I have to > have switched rings. > While 6 does have 2 as a factor in the ring of integers, it does NOT > in the ring of evens. > What I'm doing now is correcting that mistaken usage, and I'm not > surprised that it's taken some time and that many of you may have > become confused. Ah, good. I'll look forward to a cleaned-up version of the proof, then. > Here's basically my approach. I get an expression like > g = r + fc > where g, r, f, and c are algebraic integers. > Now I find out that g is not coprime to f, and I can separate off some > factor of f, which gives me > h = s + c > where h appears to be a factor of g, and s appears to be a factor of > r. > factor of f that is f, was separated off. I'm a bit confused by this step... as I understand it, if g is not coprime to f, that means that they share *some* factor, call it 'e', not that g is a multiple of f. In other words you should only be able to say: g/e = r/e + (f/e)c h = s + (f/e)c rather than dividing by the whole of f as you seem to have done above. > Now to the appears part, as in checking r, I find that r is NOT an > algebraic integer! This seems slightly weird, since r = g - fc, and if g, f, and c are all algebraic integers, r must be too. One of the other three must also turn out not be an algebraic integer. === Subject: Re: Easy proof of mathematician lies Visiting Assistant Professor at the University of Montana. > > I'm putting this bit first to highlight it: >>Note: Keith Ramsay is pointing out that he has edited my post in >>terms of placement of text, as he has moved a section. >> > [...] > | Consider that in the ring of algebraic integers, 5 has algebraic > | integer factors, and given algebraic integers r_1 and r_2, where their > | product is 5, why are you acting as if it's so difficult to comprehend > | that there must be some distribution of factors of 5? > > I'm not acting as if I'm having any kind of difficulty. I'm saying > that *you're* having difficulty in realizing that there must be some > distribution of factors of 5 is a meaninglessly vague phrase. You're > saying this as if to say, Shouldn't this be true?, and not as if > you actually had proven it was true. >>Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in >>some way between r_1 and r_2. >> It makes sense, but (assuming I know what you mean - yes you >> _are_ using the language in nonstandard ways) it's simply >> _not_ _true_. >> Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor >> of 5. Does it go with r_1 or r_2? >using standard mathematical usage, while several posters apparently >keep reading factors of 5 as multiples of 5. >The problem has to do with their misuse of mathematical terminology. It is about time you realize that you are the one who is misusing terminology. >For instance 2 is a factor of 6, but so is 3, and in fact, so is 6. >When you say, factor of, it means something that is a factor of the >given number. And similarly factors of would mean factors of the >given number. The problem is when you say x has a factor of y. This is, in standard mathematical terminology, the same as saying y is a factor of x; which is the same, in standard mathematical terminology, as y is a divisor of x and as x is a multiple of y. You do NOT mean x has a factor of y; what you mean is x has [non-unit] factors in common with y. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Easy proof of mathematician lies >> Nope. There is no claim that y is a factor as it's *provably* a >> factor. >> *sigh* Could you give me your definition of a factor, then? I suspect >> it's different from what mathematicians usually mean, because the way to >> prove y is a factor of x is to find some z such that y*z=x, with suitable >> properties for all three of them (i.e. all in the algebraic integers, >> say). If you haven't done this, how can you prove it's a factor? >I've switched from saying provably a factor, as that is problematic >to saying that it's implied to be a factor, and an analogy is in the >ring of evens where you have 2 and 6, but 6 does not have 2 as a >factor, though 6=2(3), in the ring of integers. >Here the problem is that members that should be in the ring of >algebraic integers are left out. So imagine one such member 'm', and >the algebraic integers 'c' and 'a', such that > c = am >and you'll find that in the ring of algebraic integers, 'a' is coprime >to 'c', which is the problem, as by the definition of factor, 'm' as >it's not an algebraic integer is NOT a factor of 'c' in the ring of >algebraic integers, just like 2 is NOT a factor of 6 in the ring of >evens. And, amazingly, you failed to answer the question. Here it is again, in capital letters so you don't miss it again: COULD YOU PLEASE GIVE ME YOUR DEFINITION OF A FACTOR, THEN? You know, something like: Let R be a ring, and x and y elements of R; then x IS A FACTOR OF y (in R) if and only if .... or Let R be a ring, x and y elements of R; then x HAS A FACTOR OF y (in R) if and only if ... No examples, no analogies, no talking about what you used to do and what you are doing now, no talking about why you changed the way you say things. Just give the damned definition. Don't give analogies, don't give examples, don't explain what is wrong with the usual, don't explain what is wrong with the unusual. Just give the damned definition. That is a fundamental part of mathematics (precise language, carefully defined terms). Until you learn that, you are not doing mathematics, you care doing crypto-mathematics and quackery. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Easy proof of mathematician lies [cut] >And in fact, using factors of when you mean multiples of while >common, is technically incorrect. > I have never seen anyone, here or elsewhere, use factors of > to mean multiples of. Except for that you have a good point. Ramsay said something like 6 has a factor of 3. Likewise, 15 has a factor of 3. That is, has a factor of is the same as is a multiple of. When Ramsay was discussing this, JSH conviently changed the phrase from has a factor of to the phrase is a factor of when he tried to show Ramsay he was not using standard terminology. he would find Ramsay's usage to be the standard usage. I did the search, but looked at only the first couple of hits. James would need to find a hit that said something like 15 has a factor of 10. I doubt that he would find any such usage like that. -- Bill Hale === Subject: Re: Easy proof of mathematician lies > [cut] >And in fact, using factors of when you mean multiples of while >common, is technically incorrect. > > I have never seen anyone, here or elsewhere, use factors of > to mean multiples of. Except for that you have a good point. > Ramsay said something like 6 has a factor of 3. > Likewise, 15 has a factor of 3. > That is, has a factor of is the same as is a multiple of. However, it is not. Common usage may use it that way, but expanded out, saying 15 has a factor of 3, is equivalent to saying 3 is a factor of 15, and as 3 is a factor of 3, 15 has a factor of 3. It can be a multiple of what's given but that's not forced. If the discussion were with integers then it wouldn't be a big deal for me to use what I see as slang; however, what shouldn't be lost here is that it's not that simple. A better example is, in the ring of algebraic integers, g has a factor of 5, that is 1+2i. So the context is important here. Remember that a *proof* is being discusssed while people get excited about whether or not has factors of can mean something other than is a multiple of, which should tell you something. They're trying to deceive you rather than get to the bottom of things. > When Ramsay was discussing this, JSH conviently changed the > phrase from has a factor of to the phrase is a factor of > when he tried to show Ramsay he was not using standard terminology. Really? Where? In any event a number can be said to have a factor of 5, without that meaning the number has 5 as that factor. For instance, 21 has a factor of 12, in that 3 is a factor of both 21 and 12. Now many may read that as 21 has a multiple of 12, but that's not what's stated. And in fact, you can say that 12 has a factor of 3, as 3 is itself a factor of itself, but to be precise you can say 12 is a multiple of 3. And again, if it were *integers* being discussed then I wouldn't have a problem with using 12 has a factor of 3 as meaning 12 is a multiple of 3, as that shortcut probably would be ok. However, in context, my usage fits the situation, and it seems to me that posters have a problem with my correct usage because they can't find anything wrong with the math--so they argue semantics. > he would find Ramsay's usage to be the standard usage. I did the > search, but looked at only the first couple of hits. James would > need to find a hit that said something like 15 has a factor of 10. > I doubt that he would find any such usage like that. > -- Bill Hale See what I mean? How many of you thought better of mathematicians before you saw the tricks they play? My usage is correct as any of you can demonstrate for yourselves by noticing that 12 has a factor of 21, as 3 is a factor of both. But you may also think to yourself that I should just give in to the mathematicians and posters, and play along if I want to convince them. But you see, at least some of them are mathematicians, so they are math experts! I use precision; they get upset. Rather than admit the truth--that my proofs are correct--mathematicians play word games and debate me over use of factor of because they're deceitful. === Subject: Re: Easy proof of mathematician lies Visiting Assistant Professor at the University of Montana. >> [cut] >>And in fact, using factors of when you mean multiples of while >>common, is technically incorrect. >> >> I have never seen anyone, here or elsewhere, use factors of >> to mean multiples of. Except for that you have a good point. >> Ramsay said something like 6 has a factor of 3. >> Likewise, 15 has a factor of 3. >> That is, has a factor of is the same as is a multiple of. >However, it is not. Common usage may use it that way, but expanded >out, saying 15 has a factor of 3, is equivalent to saying 3 is a >factor of 15, and as 3 is a factor of 3, 15 has a factor of 3. >It can be a multiple of what's given but that's not forced. If the >discussion were with integers then it wouldn't be a big deal for me to >use what I see as slang; however, what shouldn't be lost here is that >it's not that simple. >A better example is, in the ring of algebraic integers, g has a factor >of 5, that is 1+2i. >So the context is important here. The context is that you INSIST on using terminology in an incorrect or non-standard manner. When you say g has a factor of 5 to mean that 1+2i divides g, you are MISUSING the terminology. The CORRECT and STANDARD way of saying what you mean is to say that g and 5 have common [non-trivial] factors. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Easy proof of mathematician lies Visiting Assistant Professor at the University of Montana. >> [cut] >>And in fact, using factors of when you mean multiples of while >>common, is technically incorrect. >> >> I have never seen anyone, here or elsewhere, use factors of >> to mean multiples of. Except for that you have a good point. >> Ramsay said something like 6 has a factor of 3. >> Likewise, 15 has a factor of 3. >> That is, has a factor of is the same as is a multiple of. >However, it is not. Common usage may use it that way, but expanded >out, saying 15 has a factor of 3, is equivalent to saying 3 is a >factor of 15, and as 3 is a factor of 3, 15 has a factor of 3. >It can be a multiple of what's given but that's not forced. If the >discussion were with integers then it wouldn't be a big deal for me to >use what I see as slang; however, what shouldn't be lost here is that >it's not that simple. >A better example is, in the ring of algebraic integers, g has a factor >of 5, that is 1+2i. This is indeed a better example. It proves, beyond a shadow of a doubt, that you are misusing standard terminology; or else that you are using nonstandard terminology. If when you say g has a factor of 5, you really mean not that 5 divides g, but rather that there is an algebraic integer which divides both 5 and g, then the STANDARD AND CORRECT way of saying it is g and 5 have a common factor [in the ring of all algebraic integers]. (the bracketed part may be omitted if it is clear from context or already agreed on). If you want to further specify that this common factor is not a unit, then the STANDARD AND CORRECT way of saying it is to say: g and 5 have a common non-trivial/non-unit factor [in the ring of all algebraic integers]. In the specific case of the ring of all algebraic integers, this is equivalent to g and 5 are not coprime [in the ring of all algebraic integers] (equivalent left as an exercise exercise to the competent reader; James, you should skip it). [.snip.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Easy proof of mathematician lies > For instance, 21 has a factor of 12, in that 3 is a factor of both 21 and 12. > But Magidin, 9 does not have a factor of 12. Now that makes a lot of sense... or maybe not. === Subject: Re: Easy proof of mathematician lies > My usage is correct as any of you can demonstrate for yourselves by > noticing that 12 has a factor of 21, as 3 is a factor of both. That is some funny-ass , man. You're a humorist, right? === Subject: Re: Easy proof of mathematician lies > If when you say g has a factor of 5, you really mean not that 5 > divides g, but rather that there is an algebraic integer which divides > both 5 and g, then the STANDARD AND CORRECT way of saying it is > g and 5 have a common factor [in the ring of all algebraic > integers]. Oh, for crying out loud, is *THAT* what James has been meaning? This just shows what everyone has been saying all along, that precision and rigor in defining and using mathematical language really is important. === Subject: Re: Easy proof of mathematician lies > The problem is when you say x has a factor of y. This is, in > standard mathematical terminology, the same as saying y is a factor > of x; which is the same, in standard mathematical terminology, as y > is a divisor of x and as x is a multiple of y. I seem to have lost track of exactly what the original statements that prompted all this discussion were, but I do recall seeing things like: x has a factor of 5 which is sqrt(5). Perhaps people are just parsing this differently. One way of looking at it is as x has a value which is a factor of 5, that value being sqrt(5). It's still too ambiguous to say for certain, but that's one possible interpretation which would at least make sense and could explain why everyone insists they're using the correct definition. Of course that just emphasizes the need for precise terminology and usage... === Subject: Re: Easy proof of mathematician lies > See what I mean? How many of you thought better of mathematicians > before you saw the tricks they play? The more I read of this sort of thing, the better I think of mathematicians, and the worse I think of you. > Rather than admit the truth--that my proofs are > correct--mathematicians play word games and debate me over use of > factor of because they're deceitful. Their usage is correct; yours is incorrect. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Easy proof of mathematician lies >> If when you say g has a factor of 5, you really mean not that 5 >> divides g, but rather that there is an algebraic integer which divides >> both 5 and g, then the STANDARD AND CORRECT way of saying it is >> g and 5 have a common factor [in the ring of all algebraic >> integers]. > Oh, for crying out loud, is *THAT* what James has been meaning? > This just shows what everyone has been saying all along, that precision > and rigor in defining and using mathematical language really is important. When James says something like x has a factor of y he *sometimes* means y is a factor of x and *other* times he means x and y share a common factor. The problem is that he doesn't specify *which* meaning he's using in a given instance; indeed he appears to use them interchangeably, as if he doesn't see any difference. Sometimes he uses both meanings in the same argument, flip-flopping between them whenever it suits his convenience. That's why he doesn't like strict definitions: His so-called proofs depend on ambiguity and nebulous concepts that are clear (if anywhere) only in his own head. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Easy proof of mathematician lies [...] | See what I mean sci.skeptic? The issue here is that I'm actually | using standard mathematical usage, while several posters apparently | keep reading factors of 5 as multiples of 5. No. The problem was with the use of the phrase has a factor of. Well, if r_1 has a factor of 5 that is 1+2i, and r_2 has a factor of 5 that is 1-2i, then I'd say there it is clear there is a distribution of factors of 5. I've noted before how I talk of factors of and when I mean a specific factor I say that is, for instance, a factor of 5 that is 1+2i. You seem to think that once we agree on the meaning of is a factor of, the meaning of has a factor of is automatic. But r_1 has 1+2i doesn't mean anything. Saying r_1 has 1+2i, which is a factor of 5 is not better. r_1 has a factor of 5, which is 1+2i is equally nonstandard. It is like saying I have two parents of Bruce. If I wanted to say that we were siblings, I would need to say something like I have two parents of Bruce *as my parents*. That at least would be logical, although eccentric. You need to say something like is divisible by 1+2i, which is a factor of 5. It only makes it somewhat worse that there's already an established conventional meaning of has a factor of. [*] I have no problem with your not liking this usage. In fact people tend to avoid it when they are writing mathematics. It's kind of informal. But please don't try to replace it with another LESS standard usage. | The problem has to do with their misuse of mathematical terminology. The problem has to do with your confusing use of terminology. | For instance 2 is a factor of 6, but so is 3, and in fact, so is 6. | | When you say, factor of, it means something that is a factor of the | given number. And similarly factors of would mean factors of the | given number. Yes. | But you have these posters, like David Ullrich who is a math professor | at Oklahoma State University, who are lost with rather basic | mathematical language to the point that they *keep* arguing *after* | I've corrected them. You see their *belief* system apparently is that | as mathematicians they can't be the ones with the error, so amazingly, | they simply keep almost mindlessly repeating it. It seems to me that it's actually you who obstinately continues to argue after having been corrected. Even supposing one or both of us were arrogant, this idea that we're arrogant based on a belief that as mathematicians we are infallible is absurd. Nobody says mathematicians are infallible. They just don't agree that you're doing a very good job at finding our mistakes. I was never sure whether Pertti Lounesto was especially good at catching our mistakes either, but you have some ways to go before you're as good as he was. | You may guess that they say factors of when they mean multiples | of, but I'm using the proper terminology, in the correct way. We don't say factors of to mean multiples of. We just happen to know that what people besides yourself mean if they do say x _HAS_ a factor of Y (in it) is that x _IS_ a multiple of Y. | So when I say factors of 5, I'm NOT saying multiples of 5. For | instance, 2 and 3 are factors of 6. So when I say factors of 6 it | doesn't mean multiples of 6. And I emphasize that there is a | mathematical term multiples of which applies. I'll note that you're assuming we're making a mistake which is dumber than the kind of mistake I personally tend to expect you to make, and I think it's silly. Surely you can see by now that this is not what we're thinkinig. | And in fact, using factors of when you mean multiples of while | common, is technically incorrect. It isn't common at all! | But people like Ramsay and Ullrich are unlikely to reply to *correct* | their mistakes because the society of sci.math lets people it | considers part of the society get away with the dumbest mistakes. We're not posting corrections and apologies now because we don't have good reason to believe that we're wrong. It's not about punishment. After hundreds of fallacious proofs of Fermat, it's pretty clear that you don't care very much about getting punished for crying wolf either, do you? It's fairly difficult for anyone to get punished for dumb mistakes on usenet, regardless of status. | That's how mathematicians operate as demonstrated before your eyes. | | And that's how they can have errors in their discipline for years, and | years because they seek to by definition have a society that is | perfect, Huh? What definition? | when in fact, mathematicians are just people, and people make | mistakes. Keith Ramsay P.S. [*] It depends on the meaning of of! ;-) Factor of can mean divisor of. But it is also used in phrases like My measurement of Planck's constant was off by a factor of 2, where factor of 2 does mean a number that divides 2. It's referring to a factor, 2. Factor still means factor, but the way in which it's being associated with 2 is different. I have some sympathy with your thinking that the phrase factor *that is* 2 would be a good way to say this, but that's just not how anyone else says it. I think most often it's used when talking about formulas, to refer to factors appearing in the formula. You dropped a factor of 2 on the left hand side in this step. === Subject: Re: Easy proof of mathematician lies [...] | Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in | some way between r_1 and r_2. I would guess that you mean that there exist algebraic integers dividing 5, which also divide r_1 and r_2, except that that's a peculiar thing to ask. The obvious choice of divisors of 5 to use as divisors of r_1 and r_2 are just r_1 and r_2. So is that what do you actually mean by it? | For instance, if r_1 = r_2 = sqrt(5), you have the same factor of 5 in | both r_1 and r_2, while with r_1=5, r_2 = 1, you have all the non unit | factors of 5 with r_1. | | It hardly seems like rocket science. | | Can you explain why you as a mathematician are having difficulty in | the area Keith Ramsay? One day as I was boarding a bus, a woman asked the driver a question which sounded to me exactly like where is the outside wall? To the driver this was a bit of a problem. It wasn't a problem with the driver's knowledge of geography. He actually knew the answer to the woman's question. The question wasn't complicated. It was just a language problem. She wanted to know where the pedestrian mall in downtown was, and once the driver figured out that that's what she meant, there was no problem answering her question. That's what my problem with you is like. As soon as you can make a mathematical question clear, the answer is probably not going to be hard to find. There are tools for working this sort of thing out. [...] | The mistake is in assuming that definition of the ring of algebraic | integers does not lead to contradictions. That's absurd. There's no possible way that the *definition* of algebraic integer or ring of algebraic integers could lead to a contradiction. I've seen you clutch at straws before, but this takes the cake. [...] | I've also noted that it's odd that mathematicians would rather try to | duel with proofs than simply find an error in my proof. But, of | course, as it is a proof, there is no error, which is why I think they | try to find other means. | | That's not mathematics. That's cheating. I can demonstrate what happens if one of us simply finds an error in your proof. When you consider a factorization P(m) = (a1x + uf)(a2x + uf)(a3x + uf), you don't correctly describe what a1, a2, and a3 are. You claim they are algebraic integers, but since they vary with m this cannot be precisely correct. You talk about what values they have when m=0 and when m=1, for example. In order to satisfy such an equation, they have to be something like algebraic functions of m. People have done a fair bit of work to make good sense of the notion of algebraic function, in such a way that one can talk about the values at individual points. As m assumes different values, we can talk about the unordered triple of values of a which could be used in a factorization like that. But if we want to label one of the three as being a1, another as a2, and a third as a3, we have to make a choice somehow. It's a bit like deciding which of the roots of t^2=x is going to be called sqrt(x) and which is going to be called -sqrt(x). Part of the trouble is that there's no way to make the choice which makes it defined for all x and continuous. On the other hand, you need for there to be some kind of relationship between the values of a1 at m=0 and its values elsewhere in your argument. This mistake helps to make the rest of the argument confusing. You refer to one of the a's not being coprime to m. In what ring? The ring has to contain the a's, but they appear to be functions of m. At the bottom, and I assume someone else has pointed this out, you write out P(m) but don't distribute the factor of f^2 ;-) in all of the terms. You have (I'm taking this from the posting quoting your paper): (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3 but it should be (m^3 f^6 - 3m^2 f^4 + 3mf^2)x^3 - 3( - 1 + mf^2 )xu^2f^2 + u^3f^3 (a factor of f ;-) was left out in the last term too). This does not evaluate to 65x^3 - 12x + 1 if you substitute f=sqrt(5), u=1, m=1. It would have to be u=1/sqrt(5). But you assumed that u was an algebraic integer when you introduced it. You use that to say that f is a factor of uf, for example. You write that the values of a1, a2, and a3 don't depend on u, but that doesn't explain how it's valid to assume u is an algebraic integer and then try to apply the result to a polynomial, that you get by substituting a non-integral algebraic number for u. --- Ok, now let's see what happens. I think frustration with you over the way you respond to people who find mistakes of yours is one of the main reasons why people sometimes decide to try something different, like writing a proof that the conclusion is false. But also I would say that it's often more fun, even when dealing with our favorite posters, to go and write up my own proof of a result related to what they are writing, than it is to just react to the way they've done it. | > | If you disagree with that assessment, can you please explain why you | > | believe there should be something simple and short enough for me to | > | quote? | > In mathematics, we try to make each *mathematical* statement have its | > own well-defined meaning. (There are of course nonmathematical claims | > that we sometimes make, which aren't always precise.) If a sequence of | > well-defined mathematical statements is wrong, it's wrong because (at | > least) one of the statements is wrong. | > The key is making sure all the definitions are sound. | | As I've stated before a proof begins with a truth and proceeds by | logical statements to a conclusion that then must be true. | | Therefore it follows that definitions must be sound, or the steps will | not be logical. Well, nice to see we agree. And now that you've identified the definition of ring of algebraic integers as the culprit, really you've done essentially what I've asked-- we can quote this definition specifically. Unfortunately (or fortunately), we still have no good reason to think that the definition could create a contradiction. | > Mathematics that is vaguely enough written that it can be wrong | > without any individual statements in it being definitely wrong, is | > considered very bad. It's bad because an author hasn't defined their | > terms well enough. | | The term bad is a human term inapplicable in context as mathematics | is about truth. | | That is, a proof is neither bad nor good, it is true or false. This is naive. Realistically, one has to be concerned with the quality of a proof as well as with whether it's simply true or false. Getting rid of all errors is quite hard. So in the middle of working on a proof, the usual situation is to be working on a faulty sketch of a proof. There is a huge difference between a proof that is false, but is written *well*, and has good parts in it, so that correcting it is easier, and one that is false and written *badly*, so that it's hard even to see where mistakes might be. [...] | > The standard usage is also inconsistent with the rest of the paper. | > You write, for example, ...proving that two of the a's have a factor | > that is f at a point where you plainly have NOT shown that f is a | > factor of any of the a's, in any of the usual ways factor is | > defined. | | Hmmm...that's an interesting point. What has happened at that point | in the paper is that I've considered the constant term P(0), and the | constant term with f^2 separated off, which is P(0)/f^2 = 3x u^2 + u^3 | f, and noted that it is coprime to f. Just to be clear, why is 3x+u coprime to f? | Now I then consider g_1 at m=0, where c=g_1, and notice it has a | factor of f, Assuming u is an algebraic integer. | and then based on P(0)/f^2 being coprime to f, I have | that r + c, must have a factor of f, which proves that r must have a | factor of f that is f. I guess you mean g_1=r+c as in the lemma. What makes you think that f divides r+c? | > Elsewhere, you refer to an algebraic integer f coprime to x. This | > has no standard meaning in the context you use it, where x is a | > variable. If you meant coprime in some ring of polynomials, which is | > closest to a standard meaning, it would mean that there exists | > polynomials P and Q such that P*f+Q*x=1. But if f is an algebraic | > integer, that's possible (if and) only if f is a unit, with Q=0, and | > you also assume in the same sentence that f is a non unit. | | But f while a variable is constant, and so is x. So it is not in any | ring of polynomials. | | Unfortunately, you seem to be fixated on the *letters* as in seeing an | x you may assume that it's varying. Nope. It's constant. Says who? Taking it to be a constant is not only confusing, but inconsistent with the rest of the paper. When you introduce P(m), you remark that the new variables provide *additional* degrees of freedom. If x is a constant, along with a1, a2, and a3, what's degree of freedom did you start out with? When you refer to the factorization of P(m) into linear terms (a1x+uf)(a2x+uf)(a3x+uf), that's consistent with x being a variable. If x is a constant, there's no uniqueness in such a factorization (and not much that you can prove about the values of a1, a2, and a3). In the next step you infer that the coefficient of x^3 on the right hand side (a1a2a3) is the same as the coefficient of x^3 on the left hand side. This also is consistent with x being a variable. If two polynomials are equal for all values of x, then their corresponding coefficients have to be equal. But if x is a constant, that step is invalid. Referring to f being coprime to x isn't a tipoff that x isn't a variable, as you also consider whether something is coprime to m, and m is definitely being varied. In fact, I have no reason to believe, except your say-so, that you intended for x to be a constant until you saw my question. And it's still unclear why two of the a's are supposed to be divisible by f. | That's troubling Keith Ramsay as that's basic algebra. The problem is not with the algebra. The problem is with the writing. | The symbols | have to be understood as defined, not based on your experience of how | you may be used to seeing them. But you didn't define x. Nearly all the context implies that x is being used as a polynomial variable. What do *you* think we should do in a case like that? We can, on the one hand, try to make reasonable assumptions about things which typically are done in a certain way. Based on all the context I mention above, normally this would be because you're treating x as a polynomial variable. It seems, however, as though you want to say now that whenever I make such an assumption (which you decide to declare incorrect) it's my fault. (And a fault in algebra, too.) So really, I should assume nothing. But if I did this, you have not in your wildest dreams imagined the amount of nitpicking clarification I'd be asking you to do. You would at the very least have to say what kind of variable each variable you use is. Meanwhile, you also complain when people keep pestering you to clarify details like that. You imply they're doing it only as a distraction from the real mathematical content. You can't have it both ways. Either you agree to some reasonable reading between the lines, or you write your paper in such a way that reading between the lines isn't needed. | > | > I don't know what you think is a fair way for things to work, but the | > | > way things actually work, this kind of problem with terminology will | > | > absolutely prevent you from making any headway. | > | | > | What will prevent me from making headway is if mathematicians | > | continually lie. | > Mathematicians don't continually lie. What you keep deciding are | > lies are just disagreements with you. | | That's not true. What I've done is explain clearly and in detail. No, it's far from clear, and the detail is still poor at just the place in the argument where it needs to be best, right near the end. | In reply I find people making specious issues, like your claims about | factor when you apparently are sticking in multiple. Which again was merely a misunderstanding. | > Perhaps the biggest problem of all is that you've reached a point | > where you don't have anybody you are willing to trust, who could help | > you sort out which of the claims people are making are valid and which | > are baloney. So you tend automatically to assume that the things | > people say that seem wrong to you are baloney of some kind or another. | | How can I take people like you seriously when you have trouble with | such simple things as factor of? Because I don't have trouble with this, except with people who've decided to play Humpty Dumpty (and pay their words extra to mean what they want them to mean). If I said I had a teacher of children, and I expected people to understand me as saying that there was someone who taught both me and children, they would have a problem with that too. The problem would not be with what has a teacher means. | > This also leaves you without any very good way to learn how to improve | > your mathematical tinkering. We keep telling you different things | > which _we_ claim would help you avoid pitfalls like you keep falling | > into. But (a) you don't trust us enough to believe we've identified a | > problem with what you're doing, and (b) you don't trust us to give you | > straight advice on how to avoid it; you suspect us of just trying to | > waste your time. So you're stuck with only your own inklings of what's | > a good way to work with proofs. | | Yet I've caught you in a contradiction with yourself, where you | apparently have been inserting the word multiple when I say | something like factor of 5, so that you read multiples of 5. If | that's not what you're doing then, who knows what's going through that | brain of yours. I believe I can explain well enough. | Readers should consider what follows if they think that's too harsh. | | > | Now then in what way is it improper terminology to talk about | > | algebraic integer factors of 5? | > I didn't say it was improper to talk about algebraic integer factors | > of 5. That has a perfectly well-defined meaning. An algebraic integer | > r is a factor of 5 if it divides 5 in the algebraic integers, meaning | > that 5/r is also an algebraic integer. | | | | You're contradicting yourself. No, as I explained elsewhere. You made the leap of assuming that has a factor of should mean has a common factor with. I have a teacher of children does not mean I share a teacher with some children. 6 has a factor of 3 does not mean 6 and 3 share a factor. | Go back and read over what you said at | the top of this post. I'm tiring of this exercise in pointing out | your errors. *You* think *you're* getting tired of correcting *my* errors? Coming from you that's a laugh. Keith Ramsay === Subject: Re: Easy proof of mathematician lies >> [cut] >>And in fact, using factors of when you mean multiples of while >>common, is technically incorrect. >> >> I have never seen anyone, here or elsewhere, use factors of >> to mean multiples of. Except for that you have a good point. >> Ramsay said something like 6 has a factor of 3. >> Likewise, 15 has a factor of 3. >> That is, has a factor of is the same as is a multiple of. >However, it is not. Yes it is. >Common usage may use it that way, but expanded >out, saying 15 has a factor of 3, is equivalent to saying 3 is a >factor of 15, and as 3 is a factor of 3, 15 has a factor of 3. Huh? This is typical - your answers always seem to miss the point. We all agree that 15 has a factor of 3, 3 is a factor of 15 are both true, and say the same thing. The problem is that this is not the only way you use has a factor of - the way you use the language 15 has a factor of 25 would be true, because 5 is a factor of 25, and 15 has 5 as a factor. In _fact_ 15 has a factor of 25 is _false_, if you're using words to mean what everyone else means by them. >It can be a multiple of what's given but that's not forced. If the >discussion were with integers then it wouldn't be a big deal for me to >use what I see as slang; however, what shouldn't be lost here is that >it's not that simple. >A better example is, in the ring of algebraic integers, g has a factor >of 5, that is 1+2i. >So the context is important here. Uh, what seems to me to be important is the question that you simply ignored a _second_ time: You say that if r_1 r_2 = 5 then the factors of 5 distribute among r_1 and r_2. Let r_1 = 1+2i, r_2 = 1-2i. Note that sqrt(5) is a factor of 5 (in the algebraic integers of course). Does that mean that sqrt(5) is a factor of r_1 or of r_2? >Remember that a *proof* is being discusssed while people get excited >about whether or not has factors of can mean something other than is >a multiple of, which should tell you something. >They're trying to deceive you rather than get to the bottom of things. >> When Ramsay was discussing this, JSH conviently changed the >> phrase from has a factor of to the phrase is a factor of >> when he tried to show Ramsay he was not using standard terminology. >Really? Where? In any event a number can be said to have a factor of >5, without that meaning the number has 5 as that factor. For >instance, 21 has a factor of 12, in that 3 is a factor of both 21 and >12. No. You are _revising_ termimology here. Saying 21 has a factor of 12 and then castigating _us_ for our sloppy usage is _exactly_ the same as saying that the integers are irrational and then laughing at all the people who don't realize that's so. >Now many may read that as 21 has a multiple of 12, but that's not >what's stated. >And in fact, you can say that 12 has a factor of 3, as 3 is itself a >factor of itself, but to be precise you can say 12 is a multiple of 3. >And again, if it were *integers* being discussed then I wouldn't have >a problem with using 12 has a factor of 3 as meaning 12 is a multiple >of 3, as that shortcut probably would be ok. >However, in context, my usage fits the situation, and it seems to me >that posters have a problem with my correct usage because they can't >find anything wrong with the math--so they argue semantics. You really think that nobody's noticed that you _continue_ to ignore half of my posts here, answering (incorrectly) the part about the language but simply not replying to the part about the math? Tell me - how _does_ sqrt(5) distribute among 1+2i and 1-2i? You've laughed at people for not following this, saying it doesn't seem like rocket science. So answer the question. Ignoring questions about the math and then saying that there are no questions about the math would be funny except it's much too transparent. You can do much better. >> he would find Ramsay's usage to be the standard usage. I did the >> search, but looked at only the first couple of hits. James would >> need to find a hit that said something like 15 has a factor of 10. >> I doubt that he would find any such usage like that. >> -- Bill Hale >See what I mean? How many of you thought better of mathematicians >before you saw the tricks they play? >My usage is correct as any of you can demonstrate for yourselves by >noticing that 12 has a factor of 21, as 3 is a factor of both. But >you may also think to yourself that I should just give in to the >mathematicians and posters, and play along if I want to convince them. >But you see, at least some of them are mathematicians, so they are >math experts! I use precision; they get upset. No, you use language in totally nonstandard ways, people try to explain that what you're saying doesn't mean what you think it does and you refuse to believe them. When you insist on speaking your own private language you shouldn't be surprised when people don't believe you - they're assuming you mean what you _say_. (It's exactly this problem with has a factor of that prevented Ramsay from understanding what you meant a few posts up in this very thread. Honest.) >Rather than admit the truth--that my proofs are >correct--mathematicians play word games and debate me over use of >factor of because they're deceitful. > ************************ David C. Ullrich === Subject: Re: Easy proof of mathematician lies > [...] > | Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in > | some way between r_1 and r_2. > I would guess that you mean that there exist algebraic integers > dividing 5, which also divide r_1 and r_2, except that that's a > peculiar thing to ask. The obvious choice of divisors of 5 to use > as divisors of r_1 and r_2 are just r_1 and r_2. > So is that what do you actually mean by it? The issue that is raised at the start of the paper is a question about how one would know that given x^2 + x - 5 = (x - r_1)(x - r_2) where factors of 5 might be in r_1 and r_2, in terms of the question of whether or not either could be a unit factor of 5. For those who wonder what a unit factor is, it's like how 1 is a factor of 5, but in the ring of algebraic integers there are an infinity of numbers that are factors of 1 that are themselves not 1. For instance (1+sqrt(-3))/2 is a unit factor because it multiplies times (1-sqrt(-3))/2 to give 1, and both are algebraic integers. It turns out that mathematicians don't know how to prove whether or not r_1 and r_2 in the example I gave are unit factors. However, if you tell them that they may get upset and start showering you with a lot of mathematical statements, which if you look carefully, don't prove it, or disprove it. But mathematicians have apparently *decided* that neither r_1 nor r_2 can be unit factors of 5, though they don't know how to prove it. Unfortunately, they refuse to admit the reality, and argue using reasoning that has been shown by me to be flawed in order to claim that they can. So the preamble in my paper is highlighting that problem. I'm basically hitting them in the face with it at the start. And you'll notice how much arguing has gone on about the *preamble* and not the the main part of the paper. > | For instance, if r_1 = r_2 = sqrt(5), you have the same factor of 5 in > | both r_1 and r_2, while with r_1=5, r_2 = 1, you have all the non unit > | factors of 5 with r_1. > | It hardly seems like rocket science. > | Can you explain why you as a mathematician are having difficulty in > | the area Keith Ramsay? > One day as I was boarding a bus, a woman asked the driver a question > which sounded to me exactly like where is the outside wall? > To the driver this was a bit of a problem. It wasn't a problem with > the driver's knowledge of geography. He actually knew the answer to > the woman's question. The question wasn't complicated. It was just a > language problem. She wanted to know where the pedestrian mall in > downtown was, and once the driver figured out that that's what she > meant, there was no problem answering her question. > That's what my problem with you is like. As soon as you can make a > mathematical question clear, the answer is probably not going to be > hard to find. There are tools for working this sort of thing out. Yeah and how *long* did it take the driver? My problem with mathematicians as highlighted by the subject line of this thread is that I'm seeing sickening evidence that they are working to *hide* the truth. So Keith Ramsay, let's see if this post of yours changes that assessment in any way. You see Keith Ramsay, in the real world, people tend to try to get to the bottom of things quickly because they have a job to do, and value their time. Now I think mathematicians value their time like other people, but the twist is that I see you as working to lengthen the time in which the world is left ignorant of the truth, and that's wrong. > [...] > | The mistake is in assuming that definition of the ring of algebraic > | integers does not lead to contradictions. > That's absurd. There's no possible way that the *definition* of > algebraic integer or ring of algebraic integers could lead to a > contradiction. I've seen you clutch at straws before, but this takes > the cake. Yet I've proven my assertions mathematically, and I've presented that mathematics repeatedly. > [...] > | I've also noted that it's odd that mathematicians would rather try to > | duel with proofs than simply find an error in my proof. But, of > | course, as it is a proof, there is no error, which is why I think they > | try to find other means. > | That's not mathematics. That's cheating. > I can demonstrate what happens if one of us simply finds an error in > your proof. > When you consider a factorization P(m) = (a1x + uf)(a2x + uf)(a3x + uf), > you don't correctly describe what a1, a2, and a3 are. You claim they > are algebraic integers, but since they vary with m this cannot be > precisely correct. You talk about what values they have when m=0 and > when m=1, for example. In order to satisfy such an equation, they have > to be something like algebraic functions of m. I've talked on this issue using P(x) = x+1, in the ring of integers. Now then, your objection is like saying that it's wrong for me to say that P(x) is an integer, though the ring is declared to be integers, and x+1 will give an integer for any integer x. Now notice that P(x) = x+1 is an algebraic function of x, but it *still* gives integers for any integer x. > People have done a fair bit of work to make good sense of the notion > of algebraic function, in such a way that one can talk about the > values at individual points. As m assumes different values, we can > talk about the unordered triple of values of a which could be used > in a factorization like that. But if we want to label one of the three > as being a1, another as a2, and a third as a3, we have to make a > choice somehow. It's a bit like deciding which of the roots of t^2=x > is going to be called sqrt(x) and which is going to be called > -sqrt(x). Part of the trouble is that there's no way to make the > choice which makes it defined for all x and continuous. On the other > hand, you need for there to be some kind of relationship between the > values of a1 at m=0 and its values elsewhere in your argument. Your lengthy statement is sophistry. I can give a simple example: Consider the set {1,2,3}, and take a member m from that set. It is true that *one* and only one member of the set is even. Notice that I can get that logical conclusion without labeling. However, I can *arbitrarily* label m_1=1, m_2=2, and m_3=3, and it doesn't change that conclusion. Now I can just as easily label m_1=2, m_2=3, m_3=1, and it doesn't change that conclusion. That is, it's *still* true that only one of the m's is even!!! What I want readers to understand is that the most reasonable conclusion to draw is that mathematicians are *deliberately* lying to them. > This mistake helps to make the rest of the argument confusing. You > refer to one of the a's not being coprime to m. In what ring? The ring > has to contain the a's, but they appear to be functions of m. For a nonzero value of m, it is *proven* that *one* of the a's is coprime to f. I've explained above simply how that is possible without there being a mistake. > At the bottom, and I assume someone else has pointed this out, you > write out P(m) but don't distribute the factor of f^2 ;-) in all of > the terms. You have (I'm taking this from the posting quoting your > paper): > (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3 And for the readers, after that you'd have 65x^3 - 12x + 1. > but it should be > (m^3 f^6 - 3m^2 f^4 + 3mf^2)x^3 - 3( - 1 + mf^2 )xu^2f^2 + u^3f^3 Which is a false statement. Notice that Keith Ramsay made a change, and below he talks about *his* change as not working as if it's my problem. > (a factor of f ;-) was left out in the last term too). This does not > evaluate to 65x^3 - 12x + 1 if you substitute f=sqrt(5), u=1, m=1. It > would have to be u=1/sqrt(5). But you assumed that u was an algebraic > integer when you introduced it. You use that to say that f is a factor > of uf, for example. You write that the values of a1, a2, and a3 don't > depend on u, but that doesn't explain how it's valid to assume u is an > algebraic integer and then try to apply the result to a polynomial, > that you get by substituting a non-integral algebraic number for u. The gist of it is that the factorization for 65x^3 - 12x + 1 is independent of x, which is a trivial enough thing usually, but VERY important here. For instance, you have x^2 + 4x + 4 = (x+2)(x+2) without worrying that the factorizaton changes as x changes, as that's the point--it doesn't. What I do in my paper is use a larger more complicated expression which *includes* 65x^3 - 12x + 1, in what you might call a superset. I find a limitation, which then automatically applies. The equivalent is noting that with x=1, you have x^2 + 4x + 4 = 9. With that you *still* have the factorization, you also now have that x+2 is a factor, and indeed it is. Notice though that with the symbols gone, just looking at 9, you have *less* information, than you have with x^2 + 4x + 4. Isn't algebra grand? > --- > Ok, now let's see what happens. Indeed. > I think frustration with you over the way you respond to people who > find mistakes of yours is one of the main reasons why people sometimes > decide to try something different, like writing a proof that the > conclusion is false. But also I would say that it's often more fun, > even when dealing with our favorite posters, to go and write up my own > proof of a result related to what they are writing, than it is to just > react to the way they've done it. I want readers to consider the negative implication and condescending tone. > | > | If you disagree with that assessment, can you please explain why you > | > | believe there should be something simple and short enough for me to > | > | quote? > | | > In mathematics, we try to make each *mathematical* statement have its > | > own well-defined meaning. (There are of course nonmathematical claims > | > that we sometimes make, which aren't always precise.) If a sequence of > | > well-defined mathematical statements is wrong, it's wrong because (at > | > least) one of the statements is wrong. > | | > The key is making sure all the definitions are sound. > | As I've stated before a proof begins with a truth and proceeds by > | logical statements to a conclusion that then must be true. > | Therefore it follows that definitions must be sound, or the steps will > | not be logical. > Well, nice to see we agree. And now that you've identified the > definition of ring of algebraic integers as the culprit, really > you've done essentially what I've asked-- we can quote this definition > specifically. Unfortunately (or fortunately), we still have no good > reason to think that the definition could create a contradiction. Stated without proof. > | > Mathematics that is vaguely enough written that it can be wrong > | > without any individual statements in it being definitely wrong, is > | > considered very bad. It's bad because an author hasn't defined their > | > terms well enough. > | The term bad is a human term inapplicable in context as mathematics > | is about truth. > | That is, a proof is neither bad nor good, it is true or false. > This is naive. Realistically, one has to be concerned with the quality > of a proof as well as with whether it's simply true or false. Stated without proof. > Getting rid of all errors is quite hard. So in the middle of working > on a proof, the usual situation is to be working on a faulty sketch of > a proof. There is a huge difference between a proof that is false, > but is written *well*, and has good parts in it, so that correcting it > is easier, and one that is false and written *badly*, so that it's > hard even to see where mistakes might be. A proof cannot be false, by definition. What you might be trying to say is that when looking for a proof, a person might see the outlines of what they think is a proof. As they pursue that outline they may find pieces of it which are faulty i.e. that are false. If there is a proof there, however, it does not change. It's up to the discoverer to find it, if they can. > [...] > | > The standard usage is also inconsistent with the rest of the paper. > | > You write, for example, ...proving that two of the a's have a factor > | > that is f at a point where you plainly have NOT shown that f is a > | > factor of any of the a's, in any of the usual ways factor is > | > defined. > | Hmmm...that's an interesting point. What has happened at that point > | in the paper is that I've considered the constant term P(0), and the > | constant term with f^2 separated off, which is P(0)/f^2 = 3x u^2 + u^3 > | f, and noted that it is coprime to f. > Just to be clear, why is 3x+u coprime to f? That's just bizarre Keith Ramsay. The full expression is 3x u^2 + u^3 f and f is coprime to 3, x and u, so it's coprime to that expression, which is what I said. So why did you ask about 3x + u? > | Now I then consider g_1 at m=0, where c=g_1, and notice it has a > | factor of f, > Assuming u is an algebraic integer. The ring has already been declared to be the ring of algebraic integers. For other readers, you may be wondering why mathematicians are so clearly lying. Good question. However, consider that if they tell the truth, it makes headlines. That is, it's news big enough to dominate the pages of newspapers around the world. Now if you found you could just lie, and the world trusted you so that they let you lie for quite some time, might you not hope that you could just keep lying endlessly, and never be held accountable? > | and then based on P(0)/f^2 being coprime to f, I have > | that r + c, must have a factor of f, which proves that r must have a > | factor of f that is f. > I guess you mean g_1=r+c as in the lemma. What makes you think that > f divides r+c? It's shown in the paper. Specifically, c is constant, so as g_1 changes, there must exist r = g_1 - c, and r must change. It's determined that g_1 has a factor of f that is f at m=0. It's further determined that when m does not equal 0, separating off f^2 from P(m) leaves a constant term that is coprime to f; therefore, a factor that is f must separate off from c, and the result is coprime to f. And yes, other readers, it's a lot easier to look over that section in the paper where you have a lot more information in front of you, than figuring it out from that explanation. > | > Elsewhere, you refer to an algebraic integer f coprime to x. This > | > has no standard meaning in the context you use it, where x is a > | > variable. If you meant coprime in some ring of polynomials, which is > | > closest to a standard meaning, it would mean that there exists > | > polynomials P and Q such that P*f+Q*x=1. But if f is an algebraic > | > integer, that's possible (if and) only if f is a unit, with Q=0, and > | > you also assume in the same sentence that f is a non unit. > | But f while a variable is constant, and so is x. So it is not in any > | ring of polynomials. > | Unfortunately, you seem to be fixated on the *letters* as in seeing an > | x you may assume that it's varying. Nope. It's constant. > Says who? Taking it to be a constant is not only confusing, but > inconsistent with the rest of the paper. That is false. The key variable that changes in the paper is m. > When you introduce P(m), you remark that the new variables provide > *additional* degrees of freedom. If x is a constant, along with > a1, a2, and a3, what's degree of freedom did you start out with? The pertinent example is x^2 + 4x + 4, like from before, as you get an additional degree of freedom from the use of x here, which allows you to talk about a family of values, versus just 9. The point of algebra is that using variables allows for conclusions to be drawn about large classes of numbers rather than forcing you to check each number. Without algebra, a person can figure out that 3(3)=9. With algebra, and an additional degree of freedom, you have that (x+2)(x+2)=x^2+4x+4, which also tell you about 16, 49, and 64 along with an infinity of other numbers. > When you refer to the factorization of P(m) into linear terms > (a1x+uf)(a2x+uf)(a3x+uf), that's consistent with x being a variable. > If x is a constant, there's no uniqueness in such a factorization (and > not much that you can prove about the values of a1, a2, and a3). In > the next step you infer that the coefficient of x^3 on the right hand > side (a1a2a3) is the same as the coefficient of x^3 on the left hand > side. This also is consistent with x being a variable. If two > polynomials are equal for all values of x, then their corresponding > coefficients have to be equal. But if x is a constant, that step is > invalid. That's false. Consider again x^2 + 4x + 4 = (x+2)(x+2), as that is true whether or not you consider x variable or not. The *factorization* is what's important, and it exists without regard to whether or not you call P(x) = x^2 + 4x + 4 a polynomial, or just have x^2 + 4x + 4 with x=2. For readers that particular falsehood is an important one to consider carefully because it's one that I suggest to you is hard to explain as anything other than a willful falsehood--that is, a lie. > Referring to f being coprime to x isn't a tipoff that x isn't a > variable, as you also consider whether something is coprime to m, > and m is definitely being varied. While it is true that m is varied, it's also true that you can consider a family of values for m that are each coprime to f. That condition states that any particular m is coprime to f. > In fact, I have no reason to believe, except your say-so, that you > intended for x to be a constant until you saw my question. That's irrelevant. Remember a proof begins with a truth, and proceeds by logical steps to a conclusion which then must be true. If you have an issue it should be with the beginning, as to whether or not it is a truth, or with a logical step. > And it's still unclear why two of the a's are supposed to be > divisible by f. The conclusion is that one of the a's is coprime to f. > | That's troubling Keith Ramsay as that's basic algebra. > The problem is not with the algebra. The problem is with the writing. Stated without proof. > | The symbols > | have to be understood as defined, not based on your experience of how > | you may be used to seeing them. > But you didn't define x. Nearly all the context implies that x is > being used as a polynomial variable. What do *you* think we should do > in a case like that? You should stick to what's in the paper, and with logic. > We can, on the one hand, try to make reasonable assumptions about > things which typically are done in a certain way. Based on all the > context I mention above, normally this would be because you're > treating x as a polynomial variable. Nope. I use a factorization, and as I've pointed out, a factorization exists independent of what you mention, like with x^2 + 4x + 4 = (x+2)(x+2), as explained above. > It seems, however, as though you want to say now that whenever I make > such an assumption (which you decide to declare incorrect) it's my > fault. (And a fault in algebra, too.) So really, I should assume > nothing. You go with what's in the paper. > But if I did this, you have not in your wildest dreams imagined the > amount of nitpicking clarification I'd be asking you to do. You would > at the very least have to say what kind of variable each variable you > use is. Sounds like a lawyer tactic. > Meanwhile, you also complain when people keep pestering you to clarify > details like that. You imply they're doing it only as a distraction > from the real mathematical content. Posters have asked questions answered in the paper. Stick with the paper. > You can't have it both ways. Either you agree to some reasonable > reading between the lines, or you write your paper in such a way > that reading between the lines isn't needed. Which begs the question of your status as a math expert, as the definition of mathematician is math expert. > | > | > I don't know what you think is a fair way for things to work, but the > | > | > way things actually work, this kind of problem with terminology will > | > | > absolutely prevent you from making any headway. > | > | > | > | What will prevent me from making headway is if mathematicians > | > | continually lie. > | | > Mathematicians don't continually lie. What you keep deciding are > | > lies are just disagreements with you. > | That's not true. What I've done is explain clearly and in detail. > No, it's far from clear, and the detail is still poor at just the > place in the argument where it needs to be best, right near the end. Then why didn't you ask more questions about that section in this post? > | In reply I find people making specious issues, like your claims about > | factor when you apparently are sticking in multiple. > Which again was merely a misunderstanding. Yet several posters argued about it, and I suggest to readers that it is strong evidence of a tendency from mathematicians to care less about the truth than their own beliefs, and egos. > | > Perhaps the biggest problem of all is that you've reached a point > | > where you don't have anybody you are willing to trust, who could help > | > you sort out which of the claims people are making are valid and which > | > are baloney. So you tend automatically to assume that the things > | > people say that seem wrong to you are baloney of some kind or another. > | How can I take people like you seriously when you have trouble with > | such simple things as factor of? > Because I don't have trouble with this, except with people who've > decided to play Humpty Dumpty (and pay their words extra to mean what > they want them to mean). If I said I had a teacher of children, and I > expected people to understand me as saying that there was someone who > taught both me and children, they would have a problem with that too. > The problem would not be with what has a teacher means. My usage resolves ambiguity, and it's telling that you're *still* trying to find a way to argue about it. For readers who didn't catch the fascinating exchange I've had with several posters, a highlight is my pointing out that you can say 12 has a factor of 21 as 3 is a factor of both 12 and 21. But posters have used has a factor of to mean is a multiple of, so they'd say 21 has a factor of 3, which is correct, but can lead to ambiguity. For instance, consider the statement: x has a factor of 12. That factor could be 2, 3, or 4. Or if I move to another ring, it could be 1+i. Simply because *certain* people choose to read that as, x is a multiple of 12, does not change the reality. > | > This also leaves you without any very good way to learn how to improve > | > your mathematical tinkering. We keep telling you different things > | > which _we_ claim would help you avoid pitfalls like you keep falling > | > into. But (a) you don't trust us enough to believe we've identified a > | > problem with what you're doing, and (b) you don't trust us to give you > | > straight advice on how to avoid it; you suspect us of just trying to > | > waste your time. So you're stuck with only your own inklings of what's > | > a good way to work with proofs. > | Yet I've caught you in a contradiction with yourself, where you > | apparently have been inserting the word multiple when I say > | something like factor of 5, so that you read multiples of 5. If > | that's not what you're doing then, who knows what's going through that > | brain of yours. > I believe I can explain well enough. Then do so. > | Readers should consider what follows if they think that's too harsh. > | > | Now then in what way is it improper terminology to talk about > | > | algebraic integer factors of 5? > | | > I didn't say it was improper to talk about algebraic integer factors > | > of 5. That has a perfectly well-defined meaning. An algebraic integer > | > r is a factor of 5 if it divides 5 in the algebraic integers, meaning > | > that 5/r is also an algebraic integer. > | No, as I explained elsewhere. You made the leap of assuming that has > a factor of should mean has a common factor with. I have a teacher > of children does not mean I share a teacher with some children. > 6 has a factor of 3 does not mean 6 and 3 share a factor. And for other readers, consider dealing with several people like Keith Ramsay, who go on and on, refuse to work at getting to the bottom of things, and when corrected, they simply go off on a tangent. I'm am *tired* of having to deal with mathematicians. They're so damn irrational!!! > | Go back and read over what you said at > | the top of this post. I'm tiring of this exercise in pointing out > | your errors. > *You* think *you're* getting tired of correcting *my* errors? Coming > from you that's a laugh. > Keith Ramsay Isn't it ironic, don't you think? A little too ironic, I really do think. My tidbit from Alanis. === Subject: Re: Easy proof of mathematician lies James, go back and learn the definitions used in mathematics. It is evident that you are the only one not playing by the rules. You have to define your terms CLEARLY and stick to those definitions. Mathematicians are NOT at fault, it's you. Your ego is so big that you don't realize this. David Moran === Subject: Re: Easy proof of mathematician lies >> [...] >> | Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in >> | some way between r_1 and r_2. >> I would guess that you mean that there exist algebraic integers >> dividing 5, which also divide r_1 and r_2, except that that's a >> peculiar thing to ask. The obvious choice of divisors of 5 to use >> as divisors of r_1 and r_2 are just r_1 and r_2. >> So is that what do you actually mean by it? >The issue that is raised at the start of the paper is a question about >how one would know that given x^2 + x - 5 = (x - r_1)(x - r_2) where >factors of 5 might be in r_1 and r_2, In general a factor of 5 does not have to be in r_1 _or_ in r_2. >in terms of the question of >whether or not either could be a unit factor of 5. >For those who wonder what a unit factor is, it's like how 1 is a >factor of 5, but in the ring of algebraic integers there are an >infinity of numbers that are factors of 1 that are themselves not 1. >For instance (1+sqrt(-3))/2 is a unit factor because it multiplies >times (1-sqrt(-3))/2 to give 1, and both are algebraic integers. For anyone still wondering, a unit factor is a factor which is also a unit. _In_ the present context unit means unit in the algebraic integers, which means algebraIc integer whose reciprocal is also an algebraic integer. >It turns out that mathematicians don't know how to prove whether or >not r_1 and r_2 in the example I gave are unit factors. That's nonsense. It's _obvious_ to _me_, someone who knows nothing about such things, how to prove whether or not r_1 and r_2 in that example are unit factors! In fact they are _not_. Proof: Since r_1 and r_2 are roots of x^2 + x - 5 it's clear that 1/r_1 and 1/r_2 are roots of 1 + x - 5x^2. That's an irreducible non-monic polynomial, so 1/r_1 and 1/r_2 are not algebraic integers. This is a special case of the incredibly obvious result that an algebraic integer is a unit if and only if the _constant_ term in its minimial polynomial is plus or minus 1. > However, if >you tell them that they may get upset and start showering you with a >lot of mathematical statements, which if you look carefully, don't >prove it, or disprove it. Really? Actually if you look carefully you see that the simple argument above _does_ prove it. >[...] >For instance, consider the statement: x has a factor of 12. >That factor could be 2, 3, or 4. Or if I move to another ring, it >could be 1+i. >Simply because *certain* people choose to read that as, x is a >multiple of 12, does not change the reality. _exactly_ like the fact that certain people say that integers are rational does not change the reality that they're irrational. Hint: You're using the language _incorrectly_ here. If _everyone_ agrees that you're using the language incorrectly then you _are_ using it incorrectly - what words and phrases mean _is_ decided by majority vote. And _nobody_ but you thinks that the statement 15 has a factor of 12 is true. What you _mean_ when you say 15 has a factor of 12 is of course true. But when you say 15 has a factor of 12 people are _not_ going to know what you mean, they're going to assume you mean what anyone else would mean by the same words (making the statement obviously false.) What do you think you accomplish by inventing your own private language this way? >> | > [...] >> No, as I explained elsewhere. You made the leap of assuming that has >> a factor of should mean has a common factor with. I have a teacher >> of children does not mean I share a teacher with some children. >> 6 has a factor of 3 does not mean 6 and 3 share a factor. >And for other readers, consider dealing with several people like Keith >Ramsay, who go on and on, refuse to work at getting to the bottom of >things, and when corrected, they simply go off on a tangent. >I'm am *tired* of having to deal with mathematicians. >They're so damn irrational!!! Then why do you keep coming back for more dealings with mathematicians? >> | Go back and read over what you said at >> | the top of this post. I'm tiring of this exercise in pointing out >> | your errors. >> *You* think *you're* getting tired of correcting *my* errors? Coming >> from you that's a laugh. >> Keith Ramsay >Isn't it ironic, don't you think? >A little too ironic, I really do think. >My tidbit from Alanis. > ************************ David C. Ullrich === Subject: Re: Easy proof of mathematician lies > In the past, I've done the equivalent of saying that 6 in the ring of > evens has 2 as a factor, which is incorrect, as the implication is > that I'm talking about the ring of evens, when to be correct I have to > have switched rings. > > While 6 does have 2 as a factor in the ring of integers, it does NOT > in the ring of evens. > > What I'm doing now is correcting that mistaken usage, and I'm not > surprised that it's taken some time and that many of you may have > become confused. > Ah, good. I'll look forward to a cleaned-up version of the proof, then. I assume you mean my paper Advanced Polynomial Factorization, but as I've noted I typically used coprime so that I didn't have to use factor. When I do use factor in the paper it follows logically. > Here's basically my approach. I get an expression like > > g = r + fc > > where g, r, f, and c are algebraic integers. > > Now I find out that g is not coprime to f, and I can separate off some > factor of f, which gives me > > h = s + c > where h appears to be a factor of g, and s appears to be a factor of > r. > factor of f that is f, was separated off. > I'm a bit confused by this step... as I understand it, if g is not > coprime to f, that means that they share *some* factor, call it 'e', > not that g is a multiple of f. In other words you should only be able > to say: > g/e = r/e + (f/e)c > h = s + (f/e)c > rather than dividing by the whole of f as you seem to have done above. But then it's forced that f/e be a unit, and I explain in detail below. > Now to the appears part, as in checking r, I find that r is NOT an > algebraic integer! > This seems slightly weird, since r = g - fc, and if g, f, and c are all > algebraic integers, r must be too. One of the other three must also > turn out not be an algebraic integer. But r, g, f and c are all algebraic integers. It might help to expand out. I have a polynomial P(m) where P(m) = g_1 g_2 g_3, and P(0) is a multiple of f, as it has as a factor f^2, and also P(0)/f^2 is coprime to f. Further I have that P(m) has a factor that is f^2, which is true for all m. And I find that when m=0, g_1 has a factor that is f. Now my selection of indices is arbitrary, but I still find that there is one and only one other of the g's that has a factor that is f, when m=0, which means I've now covered the factor f^2. Notice here that basically I'm saying that two and only two of the g's can have non unit factors of f, and each has a factor that is f. The actual indices I use are arbitrary. Next I consider P(m)/f^2, with P(m)/f^2 = h_1 h_2 h_3, where the h's are factors of the g's. Now I also have for each h, something like h_1 = s_1 + d_1, where d_1 doesn't change as m changes, which is just use of my lemma, and I know that d_1 must be coprime to f, as I know that P(0)/f^2 is coprime to f, and d_1 is a factor of P(0). THAT is the basic irrefutable point. Notice that with what you have above you end up with h = s + (f/e)c, and unless f/e is a unit, you have a contradiction. === Subject: Re: Easy proof of mathematician lies [lots snipped] > Now I find out that g is not coprime to f, and I can separate off some > factor of f, which gives me > I'm a bit confused by this step... as I understand it, if g is not > coprime to f, that means that they share *some* factor, call it 'e', > not that g is a multiple of f. > And I find that when m=0, g_1 has a factor that is f. Ah, there's the trouble. In the post I was originally replying to, you indicated that g was not coprime to f, but in the followup, you clarified that g was an actual multiple of f as well, so it is reasonable to divide off the whole of f. (Note, of course, that I have no idea whether any of that is actually true, but at least the particular quibble I was replying to is settled.) === Subject: Re: Easy proof of mathematician lies Visiting Assistant Professor at the University of Montana. >> See what I mean? How many of you thought better of mathematicians >> before you saw the tricks they play? >The more I read of this sort of thing, the better I think of >mathematicians, and the worse I think of you. >> Rather than admit the truth--that my proofs are >> correct--mathematicians play word games and debate me over use of >> factor of because they're deceitful. >Their usage is correct; yours is incorrect. His usage is certainly non-standard. If he were to (a) use it consistently; and (b) give a precise definition at the beginning, or at least when confusions arise; then there would be little problems. However, as has been noted elsewhere, he does not use it consistently either: In this thread, he said: My usage is correct as any of you can demonstrate for yourselves by noticing that 12 has a factor of 21, as 3 is a factor of both. However, earlier, he had said: But Magidin, 9 does not have a factor of 12. There is nothing inherently wrong with non-standard usage. It is James's insistence on UNDEFINED and EQUIVOCAL usage which is damning. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Easy proof of mathematician lies > The issue that is raised at the start of the paper is a question about > how one would know that given x^2 + x - 5 = (x - r_1)(x - r_2) where > factors of 5 might be in r_1 and r_2, in terms of the question of > whether or not either could be a unit factor of 5. > For those who wonder what a unit factor is, it's like how 1 is a > factor of 5, but in the ring of algebraic integers there are an > infinity of numbers that are factors of 1 that are themselves not 1. > For instance (1+sqrt(-3))/2 is a unit factor because it multiplies > times (1-sqrt(-3))/2 to give 1, and both are algebraic integers. > It turns out that mathematicians don't know how to prove whether or > not r_1 and r_2 in the example I gave are unit factors. However, if > you tell them that they may get upset and start showering you with a > lot of mathematical statements, which if you look carefully, don't > prove it, or disprove it. Depends on who is doing the looking. If it is JSH, there are lots of things that he may overlook that would be easily seen by anyone who wants to see them. For each algebraic integer there is a monic polynomial with integer coefficients of minimal degree which is satisfied by that algebraic integer, and that algebraic integer is a unit in the ring of algebraic integers if and only if the constant term of that minimal polynomial is a unit in the ring of rational integers. === Subject: Re: Easy proof of mathematician lies > > In the past, I've done the equivalent of saying that 6 in the ring of > evens has 2 as a factor, which is incorrect, as the implication is > that I'm talking about the ring of evens, when to be correct I have to > have switched rings. > > While 6 does have 2 as a factor in the ring of integers, it does NOT > in the ring of evens. > > What I'm doing now is correcting that mistaken usage, and I'm not > surprised that it's taken some time and that many of you may have > become confused. > > Ah, good. I'll look forward to a cleaned-up version of the proof, then. > I assume you mean my paper Advanced Polynomial Factorization, but as > I've noted I typically used coprime so that I didn't have to use > factor. > When I do use factor in the paper it follows logically. I emphasize that important point with this minor correction to my earlier post, and the correction is below. > Here's basically my approach. I get an expression like > > g = r + fc > > where g, r, f, and c are algebraic integers. > > Now I find out that g is not coprime to f, and I can separate off some > factor of f, which gives me > > h = s + c where h appears to be a factor of g, and s appears to be a factor of > r. factor of f that is f, was separated off. > > I'm a bit confused by this step... as I understand it, if g is not > coprime to f, that means that they share *some* factor, call it 'e', > not that g is a multiple of f. In other words you should only be able > to say: > > g/e = r/e + (f/e)c > h = s + (f/e)c > > rather than dividing by the whole of f as you seem to have done above. > But then it's forced that f/e be a unit, and I explain in detail > below. > Now to the appears part, as in checking r, I find that r is NOT an > algebraic integer! > > This seems slightly weird, since r = g - fc, and if g, f, and c are all > algebraic integers, r must be too. One of the other three must also > turn out not be an algebraic integer. > But r, g, f and c are all algebraic integers. > It might help to expand out. > I have a polynomial P(m) where > P(m) = g_1 g_2 g_3, and P(0) is a multiple of f, as it has as a > factor f^2, and also P(0)/f^2 is coprime to f. > Further I have that P(m) has a factor that is f^2, which is true for > all m. > And I find that when m=0, g_1 has a factor that is f. Now my > selection of indices is arbitrary, but I still find that there is one > and only one other of the g's that has a factor that is f, when m=0, > which means I've now covered the factor f^2. Notice here that > basically I'm saying that two and only two of the g's can have non > unit factors of f, and each has a factor that is f. The actual > indices I use are arbitrary. Notice everything is still ok at m=0, but things get bizarre for nonzero m. > Next I consider P(m)/f^2, with P(m)/f^2 = h_1 h_2 h_3, where the h's > are factors of the g's. And that's where I have again accidentally used the erroneous usage because only one of the h's is an algebraic integer. That is, *reasonably* you'd suppose that the h's are all factors of the g's but only one of them can be in the ring of algebraic integers. It's a bizarre and fascinating result. That in the ring of algebraic integers is very important as the problem comes because the definition of the ring leaves this hole I've been talking about which represents an error in taught mathematics. > Now I also have for each h, something like h_1 = s_1 + d_1, where d_1 > doesn't change as m changes, which is just use of my lemma, and I know > that d_1 must be coprime to f, as I know that P(0)/f^2 is coprime to > f, and d_1 is a factor of P(0). That's still correct. > THAT is the basic irrefutable point. Notice that with what you have > above you end up with h = s + (f/e)c, and unless f/e is a unit, you > have a contradiction. And that is the irrefutable point, and no amount of talking will change that fact. My demonstration of the hole is mathematically correct and therefore irrefutable, so mathematicians should quit playing word games and work to understand rather than confuse. Remember, mathematics *currently* has this problem. So if mathematicians get to dance rather than fix it, then students will be learning bogus math that has clearly been shown to have a problem. So why would mathematicians fight the truth? I suspect it's because they've claimed perfection from errors like the one I've found. I think they're fighting for that delusion of perfection in their discipline which has now been shattered. === Subject: Re: Easy proof of mathematician lies